Evaluating function

[sagat201]

So here is the problem that I have

Let f(x) = 3x^2 + 3x + 3 and g(h) = f(2+h) - f(2) / h
Evaluate the following:
1. g(1)
2. g(0.1)
3. g(0.01)
4. g(0.001)

Notice that the values that you entered are getting closer and closer to a number we'll call L. Find L.
L = ?
This number is called the limit of g(h) as h approaches 0. It is also called the derivative of f(x) at the point when x = 2.

I tried plugging the numbers in but that didn't work. How do I start working on this?

 

[Deleted member 4993]

So here is the problem that I have

Let f(x) = 3x^2 + 3x + 3 and g(h) = f(2+h) - f(2) / h
Evaluate the following:
1. g(1)
2. g(0.1)
3. g(0.01)
4. g(0.001)

Notice that the values that you entered are getting closer and closer to a number we'll call L. Find L.
L = ?
This number is called the limit of g(h) as h approaches 0. It is also called the derivative of f(x) at the point when x = 2.

I tried plugging the numbers in but that didn't work. How do I start working on this?

That is how one should attack the problem. Good!

Please share your work in detail, with results, so that we can catch your mistake and correct it.

 

[stapel]

I tried plugging the numbers in but that didn't work.

What do you mean when you say that completing the first part of the exercise "didn't work"? What values did you get, and how did they "not work"?

Please be complete. Thank you!

 

[sagat201]

I put in 2 + h in every x so it looks like (3(2 + h)^2 + 3(2 + h) + 3) - (3(2)^2 + 3(2) + 3) / h
After distributing and combining like terms I got 3h^2 + 15h + 15 / h

Should I have done that or is there another way?

 

[Deleted member 4993]

I put in 2 + h in every x so it looks like (3(2 + h)^2 + 3(2 + h) + 3) - (3(2)^2 + 3(2) + 3) / h
After distributing and combining like terms I got 3h^2 + 15h + 15 / h

Should I have done that or is there another way?

\(\displaystyle \displaystyle g(h) \ = \ \frac{f(x+h) - f(x)}{h}\)

\(\displaystyle \displaystyle g(h) \ = \ \frac{3(x+h)^2 + 3(x+h) + 3 - 3x^2 - 3x - 3}{h}\)

\(\displaystyle \displaystyle g(h) \ = \ \frac{3(x^2 + 2xh + h^2) - 3x^2 + 3(x+h) - 3x }{h}\)

\(\displaystyle \displaystyle g(h) \ = \ \frac{6xh + 3h^2 + 3h}{h}\)

\(\displaystyle \displaystyle g(h) \ = \ \frac{h * (6x + 3h + 3)}{h}\)

\(\displaystyle \displaystyle g(h) \ = \ 6x + 3h + 3 \)

Now then, at x = 2

g(1) = 6*2+3*1+3 = 18

and so on.....

 

[sagat201]

Thank you so much for the detailed explanation! Never thought about just putting x in for 2 + h but it made the problem so much easier!